Question

Question #4c9b0

Answer 1

`"0.35 mol kg"^(-1)`

Explanation:

In order to find a solution's molality, you must know two things

• the number of moles of solute
• how many kilograms of solvent you have in the solution

You're dealing with a `2%"w/w"` sodium chloride solution, so right from the start you know that you can use the percent concentration by mass to determine how many grams of solute and how many grams of solvent you get in a random sample of this solution.

To make the calculations easier, pick a `"100-g"` sample of your solution. The given percent concentration by mass tells you that this sample will contain `"2 g"` of sodium chloride for every `"100 g"` of solution.

Use sodium chloride's molar mass to determine how many moles of solute you have in this sample

`2 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "0.03422 moles NaCl"`

Now, this solution will also contain

`m_"solution" = m_"solute" + m_"solvent"`

`m_"solvent" = "100 g" - "2 g" = "98 g water"`

This means that its molality will be - do not forget to convert the mass of water from grams to kilograms

`color(blue)(|bar(ul(color(white)(a/a)b = n_"solute"/m_"solvent"color(white)(a/a)|)))`

`b = "0.03422 moles"/(98 * 10^(-3)"kg") = "0.3492 mol kg"^(-1)`

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the percent concentration of the solution

`b = color(green)(|bar(ul(color(white)(a/a)"0.35 mol kg"^(-1)color(white)(a/a)|)))`

SIDE NOTE The molality of the solution must be the same regardless of what sample you pick as a starting point.

I highly recommend redoing the calculations for different sample of this solution - the molality will always be equal to `"0.35 kg mol"^(-1)`.